Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What are numbers $n$ such that $a^2+nb^2 = c^2$ and $na^2+b^2 = d^2$? Let $n$ and $a,b,c,d,$ be in the positive integers.
I. For the system,
$$a^2-nb^2 = c^2\\a^2+nb^2=d^2$$
then $n$ is a congruent number. The sequence starts as $n=5,6,7,13,14,15,20,21,$ and so on.
II. For,
$$a^2+b^2 = c^2\\a^2+nb^2=d^2$$
then $n$... | Note that I, II and III all contain the quadric $a^2+Nb^2=\Box$, which can be parameterized by $a=k^2-N$, $b=2k$, where $k$ is rational.
For I, $a^2-Nb^2=c^2$ gives the quartic $c^2=k^4-6Nk^2+N^2$, whilst for II, $a^2+b^2=c^2$ gives $c^2=k^4+(4-2N)k^2+N^2$. Both of these quartics have clear rational solutions $(k,c)=(0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof improvement for $(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2 + B^2$ given $(a+ib)(c+id)(e+if)(g+ih) = A + iB$ If $(a+ib)(c+id)(e+if)(g+ih) = A + iB$, prove that $(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2 + B^2$
My approach is pretty straightforward:
$$(a+ib)(c+id)(e+if)(g+ih)$$
$$((ac-bd)+i(ad+bc))(e+if)(g+ih)$$... | For complex numbers $z_1$ and $z_2$, we have $$|z_1z_2|=|z_1||z_2|$$
So,
$$(a+ib)(c+id)(e+if)(g+ih) = A + iB$$
$$|(a+ib)(c+id)(e+if)(g+ih)| = |A + iB|$$
$$|(a+ib)||(c+id)||(e+if)||(g+ih)| = |A + iB|$$
$$(\sqrt{a^2+b^2})(\sqrt{c^2+d^2})(\sqrt{e^2+f^2})(\sqrt{g^2+h^2}) = (\sqrt{A^2+B^2})$$
Thus
$$(a^2+b^2)(c^2+d^2)(e^2+f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $(1+2x)(1+x+x^2)^{n}=\sum_{r=0}^{2n+1} a_rx^r $ then find the required value If $(1+2x)(1+x+x^2)^{n}=\sum_{r=0}^{2n+1} a_rx^r $ and $(1+x+x^2)^{s}=\sum_{r=0}^{2s} b_rx^r$, then value of $\frac{\sum_{s=0}^{n}\sum_{r=0}^{2s} b_r}{\sum_{r=0}^{2n+1} \frac{a_r}{r+1}}$ will be:
(A) $2n+1$
(B) $\frac{2n+1}{2}$
(C)$\frac{n+... | $$\int_0^1(1+2x)(1+x+x^2)^{n}dx=\int_0^1\sum_{r=0}^{2n+1} a_rx^rdx = \sum_{r=0}^{2n+1} \frac{a_r}{r+1} $$
LHS integration:
$$= \frac{1}{n+1}(3^{n+1} - 1)$$
Using $(1+x+x^2)^{s}=\sum_{r=0}^{2s} b_rx^r$
Let x = 1,
$$3^s = \sum_{r=0}^{2s} b_r$$
$$\sum_{s=0}^n \sum_{r=0}^{2s} b_r = \sum_{s=0}^n 3^s = \frac{3^{n+1} - 1}{2}$... | {
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"timestamp": "2023-03-29T00:00:00",
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Question on Number theory to find sum value of n Let $S_n = n^2 + 20n + 12$, $n$ a positive integer. How do I get the sum of all possible values of $n$ for
which $S_n$ is a perfect square?
| Let $$n^2+20n+12=(n+a)^2\iff n=\dfrac{a^2-12}{2a-10}$$
So, $a$ must be even $=2b$(say)
$$\implies n=\dfrac{4b^2-12}{2(2b)-10}=\dfrac{b^2-3}{2b-5}$$
Now if integer $d$ divides $2b-5,b^2-3$
$d$ must divide $2(b^2-3)-b(2b-5)=5b-6$
$d$ must divide $2(5b-6)-5(2b-5)=13$
A necessary condition for $(2b-5)|(b^2-3)$ is $(2b-5)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $ 2^n $ divides $ 3^m-1 $ then $ 2^{n-2} $ divides $ m $ I got a difficult problem. It's kind of difficult to prove. Can you do it?
Let $ m,n\geq 3 $ be two positive integers. Prove that if $ 2^n $ divides $ 3^m -1$ then $ 2^{n-2} $ divides $ m $
Thanks :-)
| We show by induction that for any $m$ of the form $2^n q$ where $q$ is odd and $n\ge 0$, the highest power of $2$ that divides $3^{m}-1$ is $2^{n+1}$.
For the base step $n=0$, consider $3^{q}-1$. Since $3^q\equiv 3\pmod{4}$, it follows that $3^q-1$ is of the form $4t+2$, so the highest power of $2$ that divides it is... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this indefinite integral using integral substitution? So while working on some physics problem for differential equations, I landed at this weird integral $$ \int \frac 1 {\sqrt{1-\left(\frac 2x\right)}}\,dx $$
So since there is a square root, I thought I could use trig substitution, but I couldn't find a... | $u=\sqrt{1-\frac{2}{x}} \implies u^2=1-\frac{2}{x} \implies 2u du=\frac{2}{x^2} dx \implies du=\frac{1}{x^2} \frac{1}{u} dx \implies du=\frac{1}{x^2} \frac{1}{\sqrt{1-\frac{2}{x}}} dx \text{ So now back to the integral } \int \frac{1}{\sqrt{1-\frac{2}{x}}}dx=\int \frac{x^2}{x^2 \sqrt{1-\frac{2}{x}}}dx \\ \text{ now put... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$ If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$
Ideas for solution include factorizing the expression into a multiple of $x+y$ and expressing the left hand side as a sum of some perfect square expressions.
| The value of $x+y$ is not determined by the first equation. For $x=0$ we obtain $y^2-y+1=0$, so that $x+y=\frac{\pm \sqrt{-3}+1}{2}$. This is certainly not equal to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\int\frac{dx}{\cos^3x-\sin^3x}$ $\int\frac{dx}{\cos^3x-\sin^3x}$
Let $I=\int\frac{dx}{\cos^3x-\sin^3x}=\int\frac{dx}{(\cos x-\sin x)(\cos^2 x+\sin^2 x+\sin x\cos x)}$
But it does not seem to be solved further by this method,so i tried another method.
$I=\int\frac{dx}{\cos^3x-\sin^3x}=\int\frac{\csc^3 xdx}{\cot^3... | $$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x\cos x)=\dfrac{(\cos x-\sin x)\{3-(\cos x-\sin x)^2\}}2$$
Writing $\cos x-\sin x=t$ and using Partial Fraction,
$$\dfrac1{t(3-t^2)}=\dfrac1{3t}+\dfrac t{3(3-t^2)}$$
$$\implies\dfrac3{\cos^3x-\sin^3x}=\dfrac1{\cos x-\sin x}+\dfrac{\cos x-\sin x}{3-(\cos x-\sin x)^2}$$
The first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are,... | Useful identities:
$(y-z)^{3}+(z-x)^{3}+(x-y)^{3}=
3(y-z)(z-x)(x-y)$
$(y-z)^{5}+(z-x)^{5}+(x-y)^{5}=
5(y-z)(z-x)(x-y)(x^{2}+y^{2}+z^{2}-yz-zx-xy)$
$(y-z)^{7}+(z-x)^{7}+(x-y)^{7}=
7(y-z)(z-x)(x-y)(x^{2}+y^{2}+z^{2}-yz-zx-xy)^{2}$
By letting $a=y-z,b=z-x,c=x-y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 7,
"answer_id": 5
} |
$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong?
please explain why my procedure is wrong i am not able to find out??
I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions)... | Let $u=\tan \frac x2$ and kill:
\begin{align}\frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x}&=\frac{2^{\frac{2u}{1-u^2}} - 2^{\frac{2u}{1+u^2}}}{\frac{8u}{1+u^2}\arctan^2u }\\
&=(1+u^2)2^{\frac{2u}{1+u^2}}\frac{2^{\frac{4u^3}{1-u^4}} - 1}{8u\arctan^2u }\\
\end{align}
Now since $$\frac{4u^3}{1-u^4}=4\sin^3\frac x2 \times \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Make $2^8 + 2^{11} + 2^n$ a perfect square Can someone help me with this exercise? I tried to do it, but it was very hard to solve it.
Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.
It is the same thing like $4=2^2$.
| If $0\le n\le 7$, then there are no solutions. Let $n\ge 8$.
$$2^8+2^{11}+2^n=\left(2^4\right)^2\left(9+2^{n-8}\right)$$
is a square if and only if $9+2^{n-8}=m^2$ for some $m>3$, i.e. $2^{n-8}=(m+3)(m-3)$, so $m+3=2^k$ and $m-3=2^l$ for some $k>l\ge 0$. If $k\ge 4$, then $$6=2^k-2^l\ge 2^k-2^{k-1}\ge 8>6$$ contradicti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 5
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How to prove the following binomial identity How to prove that
$$\sum_{i=0}^n \binom{2i}{i} \left(\frac{1}{2}\right)^{2i} = (2n+1) \binom{2n}{n} \left(\frac{1}{2}\right)^{2n} $$
| Let
$$\begin{align}
f(i)&=(2i+1)\binom{2i}i \left(\frac 12\right)^{2i}\\
\Rightarrow f(i-1)&=\color{purple}{(2i-1)}\color{blue}{\binom{2i-2}{i-1}}\color{green}{\left(\frac 12\right)^{2i-2}}\\
&=\color{purple}{(2i-1)}\cdot\frac{\color{lightblue}{2i(2i-1)}}{\color{grey}{2i(2i-1)}}\cdot \frac {\color{grey}i}{\color{light... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\righ... | Let $t=x+3$, then
$$
((x+1)(x+2)(x+3)(x+4)(x+5))^{\frac{1}{5}}-x=(t^5-5t^3+4t)^{\frac{1}{5}}-(t-3).
$$
Using $x^5-a^5=(x-a)(x^4+ax^3+a^2x^2+a^3x+a^4)$,
\begin{align}
&(t^5-5t^3+4t)^{\frac{1}{5}}-(t-3)\\
&=\frac{t^5-5t^3+4t-(t-3)^5}{(t^5-5t^3+4t)^{\frac{4}{5}}+(t^5-5t^3+4t)^{\frac{3}{5}}(t-3)+(t^5-5t^3+4t)^{\frac{2}{5}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Tedious fraction decomposition integral $\int\frac{1}{(x^2-1)^2} \, dx$ I'm having this integral to resolve using fraction decomposition:
$$\int\frac{1}{(x^2-1)^2} \, dx$$
This gives me the following:
$$\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} = \frac{1}{(x^2-1)^2}$$
Which results in a 4 ... | You can work out the partial fraction decomposition of the integrand by repeat application of the identify:
$$\frac{1}{(x-a)(x-b)} = \frac{1}{(b-a)}\left[\frac{1}{x-a} - \frac{1}{x-b}\right]$$
This is especially useful when $a,b$ are small integers.
$$\begin{align}\frac{1}{(x^2-1)^2}
&= \left[\frac{1}{(x-1)(x+1)}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668050",
"timestamp": "2023-03-29T00:00:00",
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Find $m$ and $n$ so that $2m^2+mn+3n^2$ is a multiple of 1017
Question:
Find positive integers $m$ and $n$ so that $m+n$ is minimized and $2m^2+mn+3n^2$ is a multiple of 1017.
I found that $1017=3^2\cdot 113$ and $m=113$ and $n=4\cdot 113$ seems be the answer but I can't prove it rigorously.
| The main thing is that $$ (-23 | 113 ) = -1,$$ so that, if $2m^2 + mn + 3 n^2$ is divisible by $113,$ both $m$ and $n$ are divisible by $113.$
So your question becomes, let $2u^2 + uv + 3 v^2$ be a multiple of $9.$
Material on binary quadratic forms and the Legendre symbol are in many number theory books. Given a for... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve integral $\int \frac{\sqrt{x+1}+2}{(x+1)^2 - \sqrt{x+1}}dx$ This is how i solved it:
first i used substitution $x+1 = t^2 \Rightarrow dx=2tdt$
so integral becomes $I=\int \frac{t+2}{t^4-t}2tdt = 2\int \frac{t+2}{t^3-1}dt =2\int\frac{t+2}{(t-1)(t^2-t+1)}dt $
usic partial fraction decomposition i have:
$\frac{t+2}... |
$$\int \frac{\sqrt{x+1}+2}{(x+1)^2 - \sqrt{x+1}}dx$$
Set $s=\sqrt{x+1}$ and $ds=\frac{dx}{2\sqrt{1+x}}$
$$\int\frac{(s+2)2s}{s^4-s}ds=\int\frac{2s^2+4s}{s^4-s}ds=2\int\frac{s+2}{s^3-1}ds\overset{\text{partial fractions}}{=}2\int\frac{-s-1}{s^2+s+1}ds+2\int\frac{ds}{s-1}$$
$$=-\int\frac{2s+1}{s^2+s+1}ds-\int\frac{ds}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Probability of sum to be divisible by 7 6 fair dice are thrown simultaneously. What is the probability that the sum of the numbers appeared on dice is divisible by 7 ?
| We could approach this via markov chains and matrices.
$A=\begin{bmatrix}0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\
\frac{1}{6}&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\
\frac{1}{6}&\frac{1}{6}&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\
\frac{1}{6}&\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Square root of $\sqrt{1-4\sqrt{3}i}$ How can we find square root of the complex number
$$\sqrt{1-4\sqrt{3}i}?$$
Now here if I assume square root to be $a+ib$ i.e.
$a+ib=\sqrt{\sqrt{1-4\sqrt{3}i}}$, then after squaring both sides, how to compare real and imaginary part?
Edit: I observed
$\sqrt{1-4\sqrt{3}i}=\sqrt{4-3-... | First step:
Let $x+\mathrm i y$ a square root of $1-4\sqrt 3\mathrm i$ This means $\;x^2-y^2+2xy\mathrm i=1-4\sqrt 3\mathrm i$, whence
$$x^2-y^2=1,\quad xy=-2\sqrt 3.$$
Furthermore, $\lvert x+\mathrm i y\rvert^2=x^2+y^2=\lvert 1-4\sqrt 3\mathrm i \rvert=\sqrt{49}$.
So we have to solve the linear system (in $x^2$ and $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Quadratic expression problem involving sides of a triangle If $a,b,c$ be the sides of a triangle where $a\neq b\neq c$ and $\lambda \in R$, if roots of the equation $x^{2} + 2\lgroup a+b+c\rgroup x + 3\lambda\lgroup ab+bc+ca\rgroup =0$ are real, then what is the range of $\lambda$?
| From the quadratic formula $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ from your formula we have $$a=1,b=2(a+b+c),c=3\lambda (ab+bc+ca)$$ Subbing in $$x= \frac{-2(a+b+c) \pm \sqrt{(2(a+b+c))^2-12\lambda (ab+bc+ca)}}{2}$$ The radical is what deterimines if we will have real roots, meaning that we need $$(2(a+b+c))^2-12\lam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the inequality $\sin^8(x) + \cos^8(x) \geq \frac{1}{8}$ is true for every real number. Prove that the inequality $\sin^8(x) + \cos^8(x) \geq \frac{1}{8}$ is true for every real number.
| Slightly painful way (there may be more elegant):
Since $\sin^2+\cos^2 = 1$, you can rewrite this as
$$
(1-\cos^2 x)^4 + \cos^8 x \geq \frac{1}{8}
$$
or equivalently
$$
2 \cos^8 x - 4\cos^6 x + 6\cos^4 x - 4\cos^2 x + 1 \geq \frac{1}{8}.
$$
Setting $X=\cos^2 x$, you want first to solve the inequality
$$
2 X^4 - 4X^3 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683097",
"timestamp": "2023-03-29T00:00:00",
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$1=\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}$ $$\lim_{n \to 0} \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=1$$
this look amazingly, square root of zero is $1$.
$$s=\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}$$
$$s=\sqrt{n+s}$$
$$s=\frac{ 1\pm \sqrt{1+4n^2}}{2}$$
$$ \sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\frac{1\pm \sqrt{1+4n^2}}{2... | As $\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}\ge0$ and $\sqrt{1+4n^2}\ge1$
$\sqrt{n+\sqrt{n+\sqrt{n+ \ldots}}}=\dfrac{1+\sqrt{1+4n^2}}2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that if $p \equiv 1 \pmod 6$ then there exist integers $a$ and $b$ such that $p = a^2 + 3b^2$.
Show that if $p \equiv 1 \pmod 6$ then there exist integers $a$ and $b$ such that $p = a^2 + 3b^2$.
Let $p$ be a prime such that $p \equiv 1,3 \pmod 8$.
There exists $c \in \mathbb{Z}$ such that $c^2 \equiv −2 \pmod p$... | HINT
The hint can be proven so: $2p \equiv 2 \equiv a^2 \pmod 3$, a contradiction.
Pigeonhole Principle
Also, note that since $-3$ is a quadratic residue, this implies there exists such $a$ that $a^2 \equiv -3 \pmod p$.
By Thue's Lemma, we get that there exists such $-\sqrt{p}< x,y < \sqrt{p}$ that $x \equiv ay \pmod... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Expansion of an expression. I want to know how to expand expressions like $(x+y+z)(a+b+c)$, I currently have a problem I want to solve but I know FOIL if it is $(x+y)(a+b)$ but what do I do when it is $(x+y+z)(a+b+c)$?
| Let's create a new number $p$, which equals $y + z$. We also create $q$, which is $b + c$. Now your product becomes
$$
\begin{align}
&(x + y + z)(a + b + c) \\
&= (x + p)(a + q) \\
&= xa + xq + pa + pq \\
&= xa + x(b + c) + (y + z)a + (y + z)(b + c) \\
&= xa + xb + xc + ya + za + yb + yc + zb + zc \\
&= xa + xb + xc + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal
Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.
My attempt... | The general term of $(a+b)^n$ $$t_{r+1}={n\choose r}.a^r.b^{n-r}$$ plug in r as $6,7$ and you will get it
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.
The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.
$$y^2 − x^2 = 9, \quad y = 4; \text{ about the $x$-axis}$... | I think your limits of integration are incorrect. If you substitute $y = 4$ into $y^2 - x^2 = 9$, you find that $x = \pm \sqrt{7}$. Therefore, the two curves intersect at $x = \pm \sqrt{7}$. By washer method, we have:
$$\begin{align}
V &= \pi \int_{-\sqrt{7}}^{\sqrt{7}} (4)^2-(\sqrt{x^2+9})^2 \,dx \\
&= 2\pi \int_{0}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve fo a step by step $$\frac{81^{a}+9^{a}+1}{9^{a}+3^{a}+1}=\frac{7}{9} \Rightarrow a = ? $$
| Hint: Let $x=3^a>0$. $$\frac{81^a+9^a+1}{9^a+3^a+1}=\frac{x^4+x^2+1}{x^2+x+1}=\frac{\left(x^2+1\right)^2-x^2}{x^2+x+1}$$
$$=\frac{\left(x^2+1+x\right)\left(x^2+1-x\right)}{x^2+x+1}=x^2+1-x=\frac{7}{9}$$
So you have a quadratic equation for $x$, and then $a=\log_3(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I simplify this rational expression? I am trying to simplify $$ \frac{3(5^{k+1} - 1) + 4(3 \;\cdot\;5^{k+1})}{4}$$ to get to $$\frac{3(5^{k+2}-1)}{4} $$
I start with $$ \frac{3(5^{k+1} - 1) + 12\;\cdot\;5^{k+1}}{4}$$
$$= \frac{3\;\cdot\;5^{k+1} - 3 + 12\;\cdot\;5^{k+1}}{4} $$
then I am stuck.
| Continuing from where you left off:
$$ \frac{3\;\cdot\;5^{k+1} - 3 + 12\;\cdot\;5^{k+1}}{4} $$
Considering the numerator:
$$3\;\cdot\;5^{k+1} + 12\;\cdot\;5^{k+1} -3$$
$$5^{k+1}(3+12)-3=$$
$$5^{k+1}(15)-3=$$
$$5^{k+1}(5.3)-3=$$
$$5^{k+2}(3)-3=$$
$$3(5^{k+2}-1)$$
So your original expressing is:
$$\frac{3(5^{k+2}-1)}{4}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int \sqrt{\frac{x}{2-x}}dx$ $$\int \sqrt{\frac{x}{2-x}}dx$$
can be written as:
$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$
there is a formula that says that if we have the integral of the following type:
$$\int x^m(a+bx^n)^p dx,$$
then:
*
*If $p \in \mathbb{Z}$ we simply use binomial expansion, other... | Alternative solution - let $x=2t^2$, then
$$I=\int\sqrt{\frac{x}{2-x}}\mathrm{d}x=4\int\frac{t^2}{\sqrt{1-t^2}}\mathrm{d}t=4J$$
By parts we have
$$J=-t\sqrt{1-t^2}+\int\sqrt{1-t^2}\;\mathrm{d}t = -t\sqrt{1-t^2}+\int\frac{1-t^2}{\sqrt{1-t^2}}\;\mathrm{d}t\!=\!-t\sqrt{1-t^2}+\arcsin t-J $$
Hence
$$I=4J=2\cdot 2J =2\arcsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Verify my formula for $\sum_{n=0}^x\cos(n)=\dots$ I was wondering if I could use permutation to find the formula for $\sum_{n=0}^x\cos(n)$, and this is my work.
$$\sum_{n=0}^x\cos(n)+\cos(x+1)=\cos(0)+\sum_{n=1}^{x+1}\cos(n)$$
$$\sum_{n=0}^x\cos(n)+\cos(x+1)=1+\sum_{n=0}^x\cos(n+1)$$
$$=1+\sum_{n=0}^x\cos(n)\cos(1)-\si... | You can verify that your formula is incorrect by plugging in $x=0$.
Here is a much simpler approach. I'm assuming that $x$ is a nonnegative integer.
Multiply the desired sum by $\sin(1/2)$ and use the trig identity $\cos(a)\sin(b) = \frac{1}{2}(\sin(a+b) - \sin(a-b))$. This gives us a telescoping sum:
$$\begin{aligned}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Quotients in Ceilings and Floors How would I simplify the expression
$\lceil\frac{2x + 1}{2}\rceil - \lceil\frac{2x + 1}{4}\rceil + \lfloor\frac{2x + 1}{4}\rfloor$
I've tried writing the expression without floors or ceilings, but with no success. I also tried some casework on the parity of x.
| If $2x+1=4t+u,0<u<4$ where $t$ is any integer
$\lceil\frac{2x + 1}2\rceil - \lceil\frac{2x + 1}4\rceil + \lfloor\frac{2x + 1}4\rfloor=\lceil\dfrac u2\rceil-\lceil\dfrac u4\rceil-\lfloor\dfrac u4\rfloor$
Now $0<u<4\iff0<\dfrac u2<2$
For $0<\dfrac u2\le1,\lceil\dfrac u2\rceil=1$ and for
$1<\dfrac u2<2,\lceil\dfrac u2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How prove that: $[12\sqrt[n]{n!}]{\leq}7n+5$? How prove that: $[12\sqrt[n]{n!}]{\leq}7n+5$,$n\in N$ I know $\lim_{n\to \infty } (1+ \frac{7}{7n+5} )^{ n+1}=e$ and $\lim_{n\to \infty } \sqrt[n+1]{n+1} =1$.
| I'm assuming that $\left[x\right]
$ is the floor function. For $n=1,2
$ works. So assume that $n\geq3
$. Using the bound $$n!\leq\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}e^{1/\left(12n\right)}
$$ we have $$\left[12\sqrt[n]{n!}\right]\leq12\sqrt[n]{n!}\leq12\left(2\pi n\right)^{1/\left(2n\right)}e^{1/\left(12n^{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
$3$-adic expansion of $- \frac{9}{16}$ I get the $3$-adic expansion to be $1+1 \cdot 3+2 \cdot 3^2 +2 \cdot 3^3 + 0 \cdot 3^4+\cdots$. I'm trying to work out a pattern of the coefficients and think it is $1, 1, 2, 2, 0, 0, 1, 1, 2, 2, 0, 0,...$. To show this I need to show that $1+1 \cdot 3+2 \cdot 3^2 +2 \cdot 3^3 + 0... | I will explain the case of $3$ - adic of $\frac{21}{50}$ and leave it to you to solve for $\frac{-9}{16}$
For $\frac{21}{50}$ you have expression $\sum_{i\in \mathbb{Z}} a_i p^i$ : $0\leq a_k<p$
$$\frac{21}{50}=a_0+a_1p+a_2p^2+a_3p^3+\cdots$$
So, how do you get $a_0$ ???
$a_0$ is chosen such that
$$\frac{21}{50}=a_0+a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to prove that $a \cdot b$ is not divisible by 5 for $\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{99} + \frac{1}{100} = \frac{a}{b}$?
Let $$\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{99} + \frac{1}{100} = \frac{a}{b},$$ where $a,b$ natural numbers and $\gcd(a,b) = 1$. How to prove that $a \times b$ is not d... | If you know $\frac{1}{5k+1}+\frac{1}{5k+2}+\frac{1}{5k+3}+\frac{1}{5k+4}=\frac{a}{b}$ with $25|a$. You pretty much have it.
Let $\frac{1}{5k+1}+\frac{1}{5k+2}+\frac{1}{5k+3}+\frac{1}{5k+4}=25\frac{a_k}{b_k}$ where $5 \not \mid a_k$ and $5\not \mid b_k$.
Consider $1/1 + ... + 1/4 = 25\frac{a_0}{b_0}$ and $1/6+ .... + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find $\int \frac {\sec x}{1+ \csc x} dx $?
How to find $\int \frac {\sec x}{1+ \csc x} dx $ ?
Well it reduces to $ \int \frac {\sin x}{\cos x (1+\sin x)} dx $ .Any hints next ?
I'm looking for a short and simple method without partial fractions if possibe.
| First, we write
$$
\frac{\sec x}{1+\csc x}=\frac{\tan x}{1+\sin x}=\frac{\tan x(1-\sin x)}{1-\sin^2x}=\frac{1}{\cos^2x}\tan x-\frac{1}{\cos x}\tan^2 x.
$$
Now
$$
\int\frac{1}{\cos^2x}\tan x\,dx=\frac{1}{2}\tan^2x.
$$
For the other part we integrate by parts,
$$
\begin{aligned}
\int\frac{1}{\cos x}\tan^2x\,dx&=\int\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Trigonometric inequality in sec(x) and csc(x) How can I prove the following inequality
\begin{equation*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) \geq 3+%
\sqrt{2},~~~\forall x\in \left( 0,\frac{\pi }{2}\right) .
\end{equation*}%
I tried the following
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}... | If we set $t=\tan(x/2)$, we know that $0<t<1$ and the inequality is
$$
\left(1+\frac{1+t^2}{2t}\right)
\left(1+\frac{1+t^2}{1-t^2}\right)
\ge 3+\sqrt{2}
$$
or
$$
\frac{(1+t)^2}{2t}\frac{2}{1-t^2}\ge3+\sqrt{2}
$$
that is
$$
\frac{1+t}{t(1-t)}\ge 3+\sqrt{2}
$$
that becomes (with the condition $0<t<1$)
$$
(3+\sqrt{2})t^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Writing complex numbers in form $a+bi$ Can $\sqrt{i+\sqrt{2}}$ be expressed as $a+bi$ with $a,b \in \mathbb{R}$? In general, what kinds of expressions can be rewritten in that form?
| You can express such an expression in the form $a+ib$.
Let $$x+iy = \sqrt{i+\sqrt{2}} \\
(x+iy)^2 = (\sqrt{i+\sqrt{2}})^2 \\
x^2 - y^2 +2ixy = i+\sqrt{2} \\
$$
Now you only need to solve the equations $x^2 - y^2 = \sqrt{2}$, and $xy = \frac{1}{2}$ to get the values of $x$ and $y$.
$x = \,\,^+_-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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$a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the range of $e$.
$a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the range of $e$.
My book tells me to use tchebycheff's inequality
$$\left(\frac{a+b+c+d}{4}\right)^2\le \... | The book means the special case $x=y$, $n=4$ of the Chebyshev inequality $\overline{xy} \geq \bar{x} \bar{y}$ where $x,y$ are sequences of length $n$, both arranged in the same order (such as both increasing or both decreasing), and the bar means averaging.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How was this factoring of $1-2x-x^2$ achieved? If I factor $1-2x-x^2$ using the quadratic formula I get
$$x=\frac{2\pm \sqrt{4-4(-1)(1)}}{2(-1)}$$
$$x=\frac{2\pm \sqrt{8}}{-2}$$
$$x=-1 \pm \sqrt{2}$$
Let $\alpha = -1 +\sqrt{2}$ and $\beta=-1-\sqrt{2}$.
So $1-2x-x^2=-(x-\alpha)(x-\beta)$.
In the image below, where did ... | Note that: $\frac{1+x}{1-2x-x^2}=\frac{-(1+x)}{(\alpha-x)(\beta-x)}$ with $\alpha$ and $\beta$ as you have them in your quadratic formula part. Using partial fractions, we then get: $$=\frac{\frac{1}{2}}{\alpha-x}+\frac{\frac{1}{2}}{\beta-x}=\frac{\frac{1}{2}}{\frac{1}{\alpha}(1-\alpha x)}+\frac{\frac{1}{2}}{\frac{1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \s... | hint: Divide both sides by $\cos^2 x$ and get an quadratic equation in $\tan x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 3
} |
If the tangent at the point $P$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the major axis and minor axis at $T$ and $t$ respectively If the tangent at the point $P$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the major axis and minor axis at $T$ and $t$ respectively and $CY$ is perpendicular o... | It is the simplification of $PY$ that is not correct. However, the easy way is to note that
$$Tt\cdot PY=\overrightarrow{Tt}\cdot \overrightarrow{PC}=
\left[\matrix{-a\sec\theta\cr b\csc\theta}\right]\cdot
\left[\matrix{-a\cos\theta\cr -b\sin\theta}\right]=a^2-b^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I calculate this limit $\lim_{(x,y)\to(0,0)} \frac{xy(1-cos(x^2+y^2))}{(x^2+y^2)^{\frac{5}{2}}}$? How can I calculate this limit $$\lim_{(x,y)\to(0,0)} \dfrac{xy(1-cos(x^2+y^2))}{(x^2+y^2)^{\frac{5}{2}}}$$ at the origin?
I tried to use the substitution $x ^2 + y^2=t$ but how can I evaluate the value of $xy$? I... | Using polar coordinates $x=r\cos\theta,y=r\sin\theta$ you get $$ \begin{align*}0&\leq \lim_{(x,y)\to(0,0)} \left|\dfrac{xy(1-\cos(x^2+y^2))}{(x^2+y^2)^{\frac{5}{2}}}\right|=\lim_{r\to 0} \left|\dfrac{r^2\cos\theta\sin\theta(1-\cos(r^2))}{(r^2)^{5/2}}\right|\\&=\lim_{r\to 0}\left|\dfrac{1-\cos(r^2)}{r^3}\right|\cdot\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove $x^2+y^4=1994$ Let $x$ and $y$ positive integers with $y>3$, and $$x^2+y^4=2(x-6)^2+2(y+1)^2$$
Prove that $x^2+y^4=1994$.
I've tried finding an upper bound on the value of $x$ or $y$, but without sucess. Can anyone help me prove this problem? Note that $x^2+y^4=1994$ is the result we are trying to prove, not an a... | It follows from the condition that $x,y$ are positive integers.
Since $7^4=2401>1994$, $y$ should be less than 7.
Thus you just check 6 cases that
$y=1$,
$y=2$,
$y=3$,
$y=4$,
$y=5$,
$y=6$,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Limit similar to $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$ I want to show that
$$
\lim_{n \to \infty} \left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right) \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right)
= \lim_{n \to \infty} \prod_{k=0}^{n-1}
\left(1-\frac{n}{n^2 - k} \right)
=... | We have, by a simple comparison, $ \left( 1-\frac{n}{n^2-n} \right)^n \le \prod_{i=0}^{n-1} \left( 1-\frac{n}{n^2-i} \right) \le \left( 1-\frac{1}{n} \right)^n $. The limit of the left and right hand products as $ n \rightarrow \infty $ is $ \frac{1}{e} $, so by the Squeeze theorem, the limit of the middle product is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
Question: Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
My attempt:
$$\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$$
$$ \frac{\frac {1}{\cos(x)} - \frac{1}{\s... | Start by multiplying both sides by the denominators, so that you get
$$\sec(x)^2-\csc(x)^2 = \tan(x)^2-\cot(x)^2.$$
Now starting from the left-hand side :
\begin{align*}\sec(x)^2-\csc(x)^2 & =\frac{1}{\cos(x)^2}-\frac{1}{\sin(x)^2} \\ & = \frac{1}{\cos(x)^2}-1-\frac{1}{\sin(x)^2}+1 \\ & = \frac{1-\cos(x)^2}{\cos(x)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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How to simplify this surd: $\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}$ $$\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}} = x$$
We have to find the value of $x$.
Taking the terms to other side and squaring is increasing the power of $x$ rapidly, and it becomes unsolvable mess.
I think the answer l... | Recognize the quantity under radical sign is a perfect square, it all cancels out.
$$\sqrt{1+\frac{\sqrt{3}}{2}} = \sqrt{1+2\cdot \frac12 \cdot \frac{\sqrt{3}}{2}} = \sqrt { (\cos \pi/3 + \sin \pi/3 )^2} = (\cos \pi/3 + \sin \pi/3 ) $$
Similarly,
$$\sqrt{1-\frac{\sqrt{3}}{2}} = (\cos \pi/3 - \sin \pi/3 ) $$
Adding, ... | {
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"url": "https://math.stackexchange.com/questions/1707100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Show that $\sin x(\sin 2x+ \sin 4x +\sin 6x) = \sin 3x \sin 4x$ I got a trigonometric question in an exam and it was confused me to solve.The question contains 3 parts.
*
*Show that $\sin 2x + \sin 4x + \sin 6x = (1 + 2\cos 2x) \sin 4x$.
*By using the above show that $\sin x(\sin 2x + \sin 4x + \sin 6x) = \sin 3x \... | Once you are able to write down $\sin(3x)$ in the following way, you are good to go.
$$
\begin{align}
\sin(3x) &= \sin(x+2x)\\
& = \sin(x)\cos(2x) + \cos(x)\sin(2x)\\
&=\sin(x) \cos(2x) + \cos(x)2\sin(x)\cos(x)\\
&=\sin(x) (\cos(2x) + 2\cos^2(x)) \\
&=\sin(x) (\cos(2x) + 2\cos^2(x) -1 +1) \\
&=\sin(x) (\cos(2x) + \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find equation for mass in gravity A satellite is moving in circular motion round a planet.
From the physics we know that
$$\Sigma F_r = ma_r = \frac{GMm}{r^2}$$
So I wanted to find the equation for $M$ knowing also that
$$v = \omega r = \frac{2\pi r}{T}$$
and
$$a_r = \frac{v^2}{r}$$
Thus,
$$ma_r = \frac{GMm}{r^2}$$
$$... | The mistake lies in these steps:
$$\frac{\frac{4\pi^2r^2}{T^2}}{r} = \frac{GM}{r^2}$$
$$\frac{4\pi^2r^3}{T^2} = \frac{GM}{r^2}$$
$$\frac{4\pi^2r^5}{T^2} = GM$$
$$\frac{4\pi^2r^5G}{T^2} = M$$
Actually, it should have been:
$$\frac{\frac{4\pi^2r^2}{T^2}}{r} = \frac{GM}{r^2}$$
$$\frac{4\pi^2r}{T^2} = \frac{GM}{r^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Determine whether $p_n$ is decreasing or increasing, if $p_{n+1} = \frac{p_n}{2} + \frac{1}{p_n}$ If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.
Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24... | You have that $p_{n+1} - p_{n} = \frac{1}{p_n}-\frac{p_n}{2}$
So $p_n$ is decreasing if $ \forall n, \frac{1}{p_n} -\frac{p_n}{2} \leq 0$
And this is true if $\forall n, p_n^2 -2 \geq 0$, ie. $p_n > \sqrt{2}$ (as $p_n > 0$ )
You can prove this by induction. First, let's study the function $f(x) = \frac{x}{2}+ \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Resolve $ \frac{120}{x+y} + \frac{60}{x-y} = 6;\,\frac{80}{x+y} + \frac{100}{x-y} = 7$ I want to resolve this system of equations:
$$\begin{cases} \frac{120}{x+y} + \frac{60}{x-y} = 6 \\\frac{80}{x+y} + \frac{100}{x-y} = 7\end{cases}$$
I came to equations like
$$x - \frac{10x}{x-y} + y - \frac{10y}{x-y} = 20$$
and
$$-... | Let $u=\frac{1}{x+y}$ and $v=\frac{1}{x-y}$. Solve first for $u,v$, then you'll get equations $x+y=\frac{1}{u}$ and $x-y=\frac{1}{v}$ to solve for $x,y$.
(It's slightly easier to use $u=\frac{20}{x+y}$ and $v=\frac{20}{x-y}$ and then solve $x+y=\frac{20}{u}$ and $x-y=\frac{20}{v}$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\frac{y}{x}$ from $3x + 3y = yt = xt + 2.5x$ I need to find the ratio of
$$\frac{y}{x}$$
If given that
$$3x + 3y = yt = xt + 2.5x$$
So what I tried is:
$$t = \frac{3x + 3y}{y}$$
And then put it in the equation
$$\frac{x(3x + 3y)}{y} + 2.5x = \frac{(3x + 3y)}{y}y$$
$$\frac{x(3x + 3y)}{y} + 2.5x = 3x + 3y$$
$$\fra... | It is given that $$3x+3y=yt=xt+2.5x$$
This implies $$t=\frac{3x+3y}{y}=\frac{0.5x+3y}{x}$$
$$\Rightarrow 3x^2+2.5xy-3y^2=0$$
$$\Rightarrow -3\left(\frac{y}{x}\right)^2 +2.5\frac{y}{x}+3=0$$
$$\frac{y}{x}= \frac{-2.5 \pm \sqrt{42.25}}{-6}$$
Hence $$\frac{y}{x}=\frac{3}{2},\frac{-2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Does $p = x^2 + 9y^2$ for some $x$, $y \in \mathbb{Z}$ if and only if $p \equiv 1 \text{ mod }12$? For a prime number $p \neq 2$, $3$, does $p = x^2 + 9y^2$ for some $x$, $y \in \mathbb{Z}$ if and only if $p \equiv 1 \text{ mod }12$?
A case where this is true as to suggest plausibility: $13 = 2^2 + 9 \times 1^2$.
| A start:
A famous theorem proven by Fermat says $p = x^2 + y^2$ if and only if $p \equiv 1 \bmod 4$. Also, $x^2 \pmod 3$ only takes on the values of $0$ or $1$.
Now we have $p = x^2 + 9y^2 = x^2 + (3y)^2$.
The finish (credit to Mikhail Ivanov):
$\Rightarrow$: Let $p = 12k+1$. We know $p = x^2 + y^2$. If neither $x, y$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Inverse function of $x^2-4$ The function $h$ is defined by $$h(x)=x^2-4$$. for $$x\leq0$$
Find an expression for $h^{-1}(x)$
My attempt,
Let $h^{-1}(x)=a$
$x=h(a)$
$x=a^2-4$
$a=\sqrt{x+4}$
$h^{-1}(x)=\sqrt{x+4}$
Am I wrong? Is the answer $-\sqrt{x+4}$?
| Other answers have already correctly addressed it, but just for the sake of having another point of view, here's how I'd run through an example when I used to teach it.
We're given:
$$ h(x) = x^2 - 4, \qquad x \le 0 $$
Replace $h(x)$ with $y$:
$$ y = x^2 - 4, \qquad x \le 0 $$
Swap all $x$ and $y$, including in the res... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find all positive integers n such that $2^2 + 2^5+ 2^n$ is a perfect square. Find all positive integers n such that $2^2 + 2^5 + 2^n$ is a perfect square. Explain your answer.
| You want to solve $36+2^n=k^2\iff2^n=k^2-6^2\iff2^n=(k-6)(k+6)$
So $\exists a,b \space such\space as\space k-6=2^a, k+6=2^b\space and \space a+b=n$. You make a small table : $2^0=1,2^1=2,2^2=4,2^3=8,2^4=16..$the only values for $2^a$ and $2^b$ with an offset of 12 are $2^a=16$ and $2^b=4$. So the only solution is $n=6$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the remainder of $9^2\cdot 13\cdot 21^2$ when divided by $4$
Find the remainder of $9^2\cdot 13\cdot 21^2$ when divided by $4$
How should I approach this type of questions?
Without calculator of course
I did this:
$9^2\cdot 13\cdot 21^2=81\cdot 13\cdot 441=81\cdot 5733=464,373=33\bmod 4=1 \bmod 4$
| $(9^2\cdot13\cdot21^2)\bmod4=$
$((9\bmod4)^2\cdot(13\bmod4)\cdot(21\bmod4)^2)\bmod4=$
$(1^2\cdot1\cdot1^2)\bmod4=$
$1\bmod4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proof by induction that $1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1))}{3}$ I need to know if I am doing this right. I have to prove that
$1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1))}{3}$
So first I did the base case which would be $1$.
$1^2 = (1(2(1)-1)(2(1)+1)) / 3
1 = 3/3
1 = 1$ Which is right... | Hypothesis: $$\sum \limits_{k=1}^n (2k-1)^2 = \dfrac{n(2n-1)(2n+1)}{3} $$
For $n+1$: $$\sum \limits_{k=1}^{n+1} (2k-1)^2 = \dfrac{(n+1)(2n+1)(2n+3)}{3} = \dfrac{4n^3 - n}{3} + (2n+1)^2 = \dfrac{n(2n-1)(2n+1)}{3} + (2n+1)^2$$
which is exactly what we wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to factorize a 4th degree polynomial?
I need help to factorise the following polynomial:
$x^4 - 2x^3 + 8x^2 - 14x + 7$
The solution I need to reach is $(x-1)(x^3 - x^2 + 7x - 7)$. I need to factorize to this exactly as it is for a limit question where I cancel out the $(x-1)$ in the numerator and denominator. How... | A brute force approach is to notice that
$$\begin{align} x^4-2x^3+8x^2-14x+7
=&(x^4-x^3)-(x^3-x^2)+7(x^2-2x+1)\\
=&x^3(x-1)-x^2(x-1)+7(x-1)^2\\
=&(x-1)(x^3-x^2+7x-7)\\
=&(x-1)\left(x^2(x-1)+7(x-1)\right)\\
=&(x-1)(x-1)(x^2+7)
\end{align}$$
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove: $\tan\frac{\pi}{24}=2\sqrt{2+\sqrt{3}}-\sqrt{3}-2$
How to prove that
$$\tan\frac{\pi}{24}=2\sqrt{2+\sqrt{3}}-\sqrt{3}-2$$
I get $$\tan\frac{\pi}{24}=\sqrt\frac{2\sqrt{2}-\sqrt{3}-1}{ 2\sqrt{2}+\sqrt{3}+1}$$ but i can't transform it.
| $tan2\theta = 2tan\theta/(1-tan\theta^2)$
Let $tan\pi/12 = a, tan\pi/24 = x$, use formula, we get
$(x = -1 - \sqrt{1+a^2})/a = (-1+ \sqrt{2-\sqrt{3}})/(2-\sqrt{3}) = -1 / (2-\sqrt{3}) + 2 / \sqrt{(2-\sqrt{3})}
= -2 - \sqrt{3} + 2\sqrt{(2+\sqrt{3})}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I factorize this numerator? $$\lim_{x\to -3} \frac 1{x+3} + \frac 4{x^2+2x-3}$$
I have the solution I just need to know how i turn that into:
$$\frac {(x-1)+4}{(x+3)(x-1)}$$
I know this might be really simple but I'm not sure how to factorise the numerator.
Thanks in advance!
| First off note that $$x^2 + 2x -3 = (x+3)(x-1)$$ We can therefore rewrite your equation as $$\lim_{x \to -3} \frac 1{x+3} + \frac 4{(x+3)(x-1)} \\ \lim_{x \to -3} \frac {(x-1)}{(x+3)(x-1)} + \frac 4{(x+3)(x-1)} \\ \lim_{x \to -3} \frac {x-1 + 4}{(x+3)(x-1)} \\ \lim_{x \to -3} \frac {x+3}{(x+3)(x-1)} \\ \lim_{x \to -3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors i... | $$|\frac{a_{n+1}}{a_n}|=|\frac{n+1}{2n+1}| \to\frac{1}{2}<1$$
So by the ratio test $\sum_{n=1}^{\infty} a_n$ converges and we must have $a_n \to 0$ by the divergence test.
In fact for $n \geq 1$, $\frac{n+1}{2n+1}$ is decreasing so,
$$\frac{a_{n+1}}{a_n}:=f(n) \leq \frac{1+1}{2(1)+1}=\frac{2}{3}$$
And because the posit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
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Finding area bounded by two graphs of functions I know that someone posted this same question before here, because i found it, but i couldn't find the answer about this example that i was looking for, so i will post everything i did so far:
I have to find area bounded by graphics of functions
$y_1=x\sqrt{4x-x^2}$
$y_2... | Your integral is incorrect, you could check that on wolfram alpha.
For
$$\int_0^1 (1-x)\sqrt{4x-x^2}dx$$
Because $4x-x^2=4-(x-2)^2$
Let $x-2=2\sin t$
$$
\begin{align}
I
&=\int_{-\pi/2}^{-\pi/6} (-1-2\sin t)(2\cos t)(2\cos t)dt \\
&=-2\int_{-\pi/2}^{-\pi/6} (2\cos^2t -1+1)dt + 8\int_{-\pi/2}^{-\pi/3} \cos^2 t d(\cos t) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
integrate $\frac{\sqrt{\arctan\frac{1}{x}}}{1+x^2}$ I was trying to solve $\int \frac{\sqrt{\arctan\frac{1}{x}}}{1+x^2} dx$. I think I have the first step right, which is $\int \sqrt{\arctan\frac{1}{x}} d(\arctan x) $, but I'm not sure how to proceed. I considered integration by parts, but it seems to make the new inte... | $$
u = \arctan\left(\frac{1}{x}\right) \rightarrow du = \frac{1}{1+\frac{1}{x^2}} \cdot \frac{-1}{x^2} dx = \frac{-1}{1+x^2} dx
$$
$$
\int \frac{\sqrt{\arctan\frac{1}{x}}}{1+x^2} dx = \int -\sqrt{u} du = -\frac{2}{3}u^{3/2} + C = -\frac{2}{3}\left(\arctan\left(\frac{1}{x}\right)\right)^{3/2} + C
$$
So:
$$
\int \frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to solve $\int \dfrac{x^5\ln\left(\frac{x+1}{1-x}\right)}{\sqrt{1-x^2}} dx$ Consider the integral
$$\int \dfrac{x^5\ln\left(\frac{x+1}{1-x}\right)}{\sqrt{1-x^2}}dx$$
How to start integrating?
Any hint would be appreciated.
| $$
\int \dfrac{x^5\ln\left(\frac{x+1}{1-x}\right)}{\sqrt{1-x^2}}dx
$$
$$
u = \ln\frac{x+1}{1-x} \quad dv = \frac{x^5 dx}{\sqrt{1-x^2}}
$$
$$
v = \int \frac{x^5 dx}{\sqrt{1-x^2}} \quad x = \sin\theta \quad v = \int \sin^5\theta d\theta = -\int (1 - 2\cos^2\theta + \cos^4\theta ) d\cos\theta = \\ - \cos\theta + 2/3\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Taking inverse Fourier transform of $\frac{\sin^2(\pi s)}{(\pi s)^2}$ How do I show that
$$\int_{-\infty}^\infty \frac{\sin^2(\pi s)}{(\pi s)^2} e^{2\pi isx} \, ds = \begin{cases} 1+x & \text{if }-1 \le x \le 0 \\ 1-x & \text{if }0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}$$
I know that $\sin^2(\pi s)=\frac{1-\co... | We can prove this formally without using the given identity; just using the various Fourier transform identities.
First consider, \begin{align*} \mathcal F^{-1}(\sin(\pi s)) &= \frac{1}{2i} \int_{\mathbb R} (e^{i\pi s} - e^{-i\pi s})e^{2\pi i xs} ds \\ &= \frac{1}{2i} \int_{\mathbb R} e^{2i\pi s\left(x + \tfrac 1 2 \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What is $2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$ equal to? I came across this question while doing my homework:
$$\Large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty=?$$
$$\small\text{OR}$$
$$\large\prod\limits_{x=1}^{\... | Olympiad Tricks
\begin{align*}
y &=
1^{\frac{1}{2}} \times 2^{\frac{1}{4}} \times 4^{\frac{1}{8}} \times
\ldots \times 2^{\frac{n-1}{2^{n}}} \times \ldots \\[3pt]
y^{2} &=
2^{\frac{1}{2}} \times 4^{\frac{1}{4}} \times 8^{\frac{1}{8}} \times
\ldots \times 2^{\frac{n}{2^{n}}} \times \ldots \\[3pt]
\frac{y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Knowing binomial expansion of $(1+kx)^n$ shortcut for $(a+kx)^n$ If I've calculated the binomial expansion of $(1+kx)^n$ (where $k$ and $n$ are known) up to some term, is there then a shortcut for calculating $(a+kx)^n$?
Example: Given that the binomial expansion of $(1+6x)^{\frac23}$ up to and including the term in $... |
Hint: Let's assume you already know $$f(x)=(1+kx)^n$$
Since
\begin{align*}
(a+kx)^n=a^n\left(1+k\frac{x}{a}\right)^n\tag{1}
\end{align*}
you can calculate
\begin{align*}
(a+kx)^n=a^nf\left(\frac{x}{a}\right)\tag{2}
\end{align*}
Comment: Some details due to OPs comment
*
*In (1) we use
\begin{align*}
(a+kx)^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Suppose that a sequence is Cesaro summable. Prove.... Suppose that a sequence $a_{n}$ is Cesaro summable. Prove that
$$\lim_{n \to \infty }\frac{a_{n}}{n}=0$$
| Let $S_m$ denote the $m$th partial sum of the series:
$$S_m = \sum_{k=1}^{m}a_k$$
Let $T_n$ denote the average of the first $n$ partial sums:
$$T_n = \frac{1}{n}\sum_{m=1}^{n}S_m$$
Since $\sum a_n$ is Cesaro summable, $T_n$ converges to some limit $L$, so $T_n - T_{n-1} \to L - L = 0$ as $n \to \infty$.
Now,
$$\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.
If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.
I feel that the pr... | A theoretical and easy proof for $n$ odd.
According to a theorem of Gauss the positive integer $n$ is a sum of three squares if and only if $n$ is not of the form $4^a(8b-1)$; $a,b\in \mathbb Z$.
If $n^2=8B-1$ then $n^2\equiv -1 \pmod8$. This is not possible because $-1$ is not a square modulo $8$.
Thus the odd integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Show that X has a Chi Square Distribution with 4 degrees of freedom Let $X$ be a random variable such that $E(X^n) = (n+1)!\cdot2^n$ for $n = 1,2,3,\ldots$.
Show that $X$ has a Chi Square distribution with 4 degrees of freedom.
I am not sure how to begin this. Can I find the pdf from moments?
| A Chi Square distribution with 4 degrees of freedom has a pdf given by
$$ f_X(x) = \frac{x e^{-x/2}}{4}$$ and mgf $$M_X(t)= (1-2t)^{-2}.$$
Calculating the moments we have,
$M^{(1)}_X(0) = (-2) \cdot (1)^{-3} \cdot (-2) = 2 \cdot 2$
$M^{(2)}_X(0) = (-2) (-3) \cdot (1)^{-4} \cdot (-2)(-2) = 3 \cdot 2 \cdot 2^2$
$M^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An inequality involving $\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$
$$\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$$
Let $(x, y, z)$ be non-negative real numbers such that $x^2+y^2+z^2=2(xy+yz+zx)$.
Question: Find the maximum value of the expression above.
My attempt:
Since $(x,y,z)$ can be non-negative, we can take $x=0$... | Hint
As we know:
$$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz$$
and $$x^2+y^2+z^2=2(xy+yz+zx)$$
So,
$$x^3+y^3+z^3=\frac{1}{2}(x+y+z)(x^2+y^2+z^2)+3xyz$$
Substituting this in the expression gives:
$$\frac{\frac{1}{2}(x+y+z)(x^2+y^2+z^2)+3xyz}{(x+y+z)(x^2+y^2+z^2)}$$
$$=\frac{1}{2}+\frac{3xyz}{(x+y+z)(x^2+y^2+z^2)}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-... | HINT: Split your last sum into the sum of two arithmetic progressions, each of length $n$.
An alternative is to calculate a few values of the sum, guess a closed form, and then prove the closed form. For $n=1,2,3$ the sum in question is $20,72,156$, respectively. Note that $20=4\cdot5$, $72=8\cdot9$, and $156=12\cdot 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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A consequence of the convexity of $ f(x) = x \log x $ I verified that $f:\mathbb{R_{+}^{*}} \rightarrow \mathbb{R}, f(x) = x \log x $ is convex, since it is twice differentiable and $f''(x) = \frac{1}{x}$ is positive for the domain. But my teacher asked to verify the following consequence:
$ \forall a,b,x,y > 0 ,$ $... | Consider $x/a=X$ and $y/b=Y$. Then
$$
\frac{x+y}{a+b}=\frac{aX+bY}{a+b}=\frac{a}{a+b}X+\frac{b}{a+b}Y
$$
so we can set $t=b/(a+b)$ and $1-t=a/(a+b)$. Then
$$
\frac{x+y}{a+b}=(1-t)X+tY
$$
and, by convexity,
$$
\frac{x+y}{a+b}\log\frac{x+y}{a+b}\le
(1-t)f(X)+tf(Y)=
\frac{a}{a+b}\frac{x}{a}\log\frac{x}{a}+\frac{b}{a+b}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the integers
Find all integers $a,b,c$ with $1<a<b<c$ such that $(a-1)(b-1)(c-1)$
is a divisor of $abc-1$.
I cannot understand how to solve this. I would appreciate any help.
| Assume $1<a<b<c$ and $(a-1)(b-1)(c-1)\mid abc-1$ and let $d=\frac{abc-1}{(a-1)(b-1)(c-1)}$.
Then from $$(a-1)(b-1)(c-1)<ab(c-1)=abc-ab<abc-1 $$
we see that $d\ge2$.
Assume $a\ge 4$. Then $b\ge 5$, $c\ge 6$ and
$$abc-1=d(a-1)(b-1)(c-1)\ge d\cdot \frac34a\cdot\frac 45b\cdot \frac56c=\frac d2abc \ge abc,$$
contradiction!... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find areas of two triangles in a large triangle Can anyone help me with this problem? I can find area of $\triangle BDE$ easily, but not $\triangle EFA$.
Problem: As shown in the figure, the area of $\triangle ABC$ is $5$, $AE=ED$, and $BD =\frac{2}{3}BC$. Find the areas of triangles $\triangle BDE$ and $\triangle EFA... | Denote $(MNP)$ the area of $\triangle MNP$. Thus we have:
$(EFA) = \dfrac{(DFA)}{2}= \dfrac{(DCA)-(DCF)}{2}$. And $(DCA) = \dfrac{(ABC)}{3} = \dfrac{5}{3}$. Also, $(BDE) = \dfrac{(BDA)}{2}= \dfrac{1}{2}\cdot \dfrac{2(ABC)}{3}= \dfrac{5}{3}$. Plus: $\dfrac{(DCF)}{(BDF)} = \dfrac{DC}{BD} = \dfrac{1}{2}\Rightarrow \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof using deductive reasoning I need to deductively prove that the sum of cubes of $3$ consecutive natural numbers is divisible by $9$. I can prove deductively that they are divisible by $3$ but so far any combination I choose fails to prove the divisibility by $9$. As far as I can see. This is a high school question... | It may be more useful to have the center number be $x$, and the two numbers to either side be $x-1$ and $x+1$. You have then the sum of three consecutive cubes is $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x=3x(x^2+2)$.
Now, note that either $x$ is a multiple of $3$ or $(x^2+2)$ is a multiple of three.
Proof: $x=3k\Rightarrow x\eq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $n - \frac{_{2}^{n}\textrm{C}}{2} + \frac{_{3}^{n}\textrm{C}}{3} - ...= 1 + \frac{1}{2} +...+ \frac{1}{n}$ Prove that $n - \frac{_{2}^{n}\textrm{C}}{2} + \frac{_{3}^{n}\textrm{C}}{3} - ... (-1)^{n+1}\frac{_{n}^{n}\textrm{C}}{n} = 1 + \frac{1}{2} + \frac{1}{3} +...+ \frac{1}{n}$
I am not able to prove this. Plea... | @Olivier Oloa's proof is so brilliant! Even though,
we can still prove this by induction on $n$.
For $n=1$, then the result is clear. Now, suppose that the result holds for some $n\in\mathbb{N}$, then for $n+1$, we have
\begin{align}
\sum_{k=1}^{n+1}\frac{(-1)^{k+1}}{k}{n+1\choose k}
&=\sum_{k=1}^n\frac{(-1)^{k+1}}{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Roots of a Quartic (Vieta's Formulas)
Question: The quartic polynomial $x^4 −8x^3 + 19x^2 +kx+ 2$ has four distinct real roots denoted
$a, b, c,d$ in order from smallest to largest. If $a + d = b + c$ then
(a) Show that $a + d = b + c = 4$.
(b) Show that $abcd = 2$ and $ad + bc = 3$.
(c) Find $ad$ and $bc.$
(d) F... | You already have lots of equations, some of them non-linear, from where hopefully the coefficients will be found:
$$\begin{align*}&a+d=4=b+c\\&ad=2\,,\,\,bc=1\end{align*}$$
so for example (as clearly all coefficients are non-zero):
$$c=\frac1b\implies 4=b+c=b+\frac1b\implies b^2-4b+1=0\implies b_{1,2}=2\pm\sqrt3\implie... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence and limit of $\sum_{j=1}^n\sin\left(\frac{j-1}n\right)\left\{\cos\left(\frac jn\right)-\cos\left(\frac{j-1}n\right)\right\}$ The title says it all - I'm trying to find a way of proving the convergence and evaluating the limit of $a_n=\sum_{j=1}^n\sin\left(\frac{j-1}n\right)\left\{\cos\left(\frac jn\right)-\... | From trigonometric identities:
$$
\sum_{j=1}^{n}\sin\left(\frac{j-1}{n}\right)\left[\cos\left(\frac{j}{n}\right)-\cos\left(\frac{j-1}{n}\right)\right]
=-2\sin\frac{1}{2n}\sum_{j=1}^{n}\sin\left(\frac{j-1}{n}\right)\sin\left(\frac{2j-1}{2n}\right).
$$
Now, we also can rewrite
$$
\sin\frac{2j-1}{2n} = \sin\left(\frac{j-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to square both the sides of an equation? Question: $x^2 \sqrt{(x + 3)} = (x + 3)^{3/2}$
My solution: $x^4 (x + 3) = (x + 3)^3$
$=> (x + 3)^2 = x^4$
$=> (x + 3) = x^2$
$=> x^2 -x - 3 = 0$
$=> x = (1 \pm \sqrt{1 + 12})/2$
I understand that you can't really square on both the sides like I did in the first step, howe... | You have to specify the equation domain of validity: $D=[-3,+\infty)$.
On this domain,
\begin{align*}x^2 \sqrt{(x + 3)} = (x + 3)^{3/2}&\iff x^4(x + 3)= (x + 3)^3\iff(x+3)\Bigl[x^4-(x+3)^2\Bigr]=0 \\
&\iff(x+3)(x^2-x-3)(x^2+x+3)=0 \\&\iff(x+3)(x^2-x-3)=0 \qquad\text{on }\,D
\end{align*}
Now the second equation: $\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Olympiad Inequality AM-GM (easy) Prove that
$(1 + x + y)^2 + (1 + y + z)^2 + (1 + z + x)^2 ≤ 3(x + y + z)^2$, with equality if and only if $x = y = z = 1$ ($xyz \ge 1$) ($x,y,z$ positive reals)
This simplifies to
$x^2 + y^2 + z^2 + 4(xy + yz + zx) - 4(x + y + z) \geq 3$
I am looking for a proof using AM-GM preferably.... | $$x^2 + y^2 + z^2 + 4(xy + yz + zx) - 4(x + y + z) =(x+y+z-2)^2-4+2(xy + yz + zx)$$
Now, by AM-GM:
$$x+y+z\ge3\sqrt[3]{xyz}\ge3\tag{1}$$
$$xy+yz+zx\ge3\sqrt[3]{x^2y^2z^2}\ge3\tag{2}$$
which pretty much settles the matter:
$$\underbrace{(x+y+z-2)^2}_{\ge1}-4+2(\;\underbrace{xy + yz + zx}_{\ge3}\;)\ge1-4+6=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} = \tan^{-1}\frac{n}{n+2}$ Prove that
$$ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} = \tan^{-1}\frac{n}{n+2}$$
I have been trying to solve it step by like $ \tan^{-1}\frac{1}{3} + \tan^{-1}\f... | HINT: $$\sum_{r=1}^n \tan^{-1}\frac{1}{r^2+r+1}$$
$$=\sum_{r=1}^n \tan^{-1}\frac{(r+1)-r}{1+r(r+1)}$$
$$=\sum_{r=1}^n [\tan^{-1} (r+1) -\tan^{-1} r]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of $\frac{n^4}{\binom{4n}{4}}$ as $n \rightarrow \infty$ $\frac{n^4}{\binom{4n}{4}}$
$= \frac{n^4 4! (4n-4)!}{(4n)!}$
$= \frac{24n^4}{(4n-1)(4n-2)(4n-3)}$
$\rightarrow \infty$ as $n \rightarrow \infty$
However, the answer key says that
$\frac{n^4}{\binom{4n}{4}}$
$= \frac{6n^3}{(4n-1)(4n-2)(4n-3)}$ this ... | You have, as $n \to \infty$,
$$
\frac{n^4}{\binom{4n}{4}}=\frac{n^4 4! (4n-4)!}{(4n)!}=\frac{\color{red}{24n^4}}{\color{red}{4n}(4n-1)(4n-2)(4n-3)}=\frac{\color{red}{6n^3}}{(4n-1)(4n-2)(4n-3)}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Evaluating $\lim\limits_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x}$. Is Wolfram wrong or is it me? What am I doing wrong?
My attempt
$$\begin{align}
\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} &= \lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} \cdot \frac{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}}{\sqrt{x^... | Your error is when you divided the numerator by $x$ but you divided the denominator by $\sqrt{x^2}$. Note that when $x$ is negative, $\sqrt{x^2}$ is positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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$LDL^t$ Factorization Algorithm to find a factorization of the form A For $$
\begin{pmatrix}
4 & 2 & 2 \\
2 & 6 & 2 \\
2 & 2 & 5 \\
\end{pmatrix}
$$ I found that $$
L=\begin{pmatrix}
1 & 0 & 0 \\
1/2 & 1 & 0 \\
1/2 & 1/5 & 1 \\
\end{pmatrix... | Using the algorithm located at this site, you should get:
$$L = \left(
\begin{array}{ccc}
1 & 0 & 0 \\
\dfrac{1}{2} & 1 & 0 \\
\dfrac{1}{2} & \dfrac{1}{5} & 1 \\
\end{array}
\right)$$
$$D = \left(
\begin{array}{ccc}
4 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \dfrac{19}{5} \\
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving a differential equation: $y^4 \left(\frac{dy}{dx}\right)^4 = (y^2-1)^2$ How would one go ahead solving the following for these conditions: (i) passes through $(0,\frac{\sqrt{3}}{2})$ and (ii) $(0,\frac{\sqrt{5}}{2})$ :
$$y^4 \left(\frac{dy}{dx}\right)^4 = (y^2-1)^2$$
There is something going on with the signs h... | $$(yy')^4=(y^2-1)^2$$
$$(yy')^2=\pm(y^2-1)$$
$$yy'=\pm\sqrt{\pm(y^2-1)}$$
$Y=y^2$
$$\frac{1}{2}Y'=\pm\sqrt{\pm(Y-1)}$$
$$\frac{dY}{\pm 2\sqrt{\pm(Y-1)}}=dx$$
$$\pm\sqrt{\pm(Y-1)}=x+c$$
$$Y=1\pm(x+c)^2$$
$$y=\pm\sqrt{1\pm(x+c)^2}$$
Condition $y(0)=\frac{\sqrt{3}}{2}=+\sqrt{1- c^2} \qquad\to\qquad c=\pm\frac{1}{2}$
Condi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $1.0000...$ ( $1$ with infinite zeros) greater than $1.0$? Given that $0.3333...$ is greater than $0.3$ and similarly $0.777...$ is greater than $0.7$, does it follow that the sum of $0.33...$ and $0.77...$ is greater than sum of $0.3$ and $0.7$?
| This is clearer to see when using fraction notation:
$$0.\dot3 = \frac{1}{3},\,0.\dot7 = \frac{7}{9},\,0.3 = \frac{3}{10}, 0.7 = \frac{7}{10}$$
So:
$$0.\dot3 + 0.\dot7 = \frac{1}{3} + \frac{7}{9} = \frac{10}{9} = 1.\dot{1} > 1\\0.3+ 0.7= \frac{3}{10} + \frac{7}{10} = \frac{10}{10} = 1$$
Not only is it greater, it doesn... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What's the value of $x$ in the following equation?
So this is how I approached this question, the above equations could be simplified to :
$$a = \frac{4(b+c)}{b+c+4}\tag{1!}$$
$$b = \frac{10(a+c)}{a+c+10}\tag{2}$$
$$c=\frac{56(a+b)}{a+b+56}\tag{3}$$
From above, we can deduce that $4 > a$ since $\frac{(b+c)}{b+c+4} < 1... | You seem to be assuming through out that $a$,$b$,$c$ are positive integers (e.g. in your deduction that $a < 4$), which doesn't appear to be in the problem statement.
Given they're all integers, I would clear denominators so you don't have to deal with fractions. We then notice that the last equation becomes $x(a+b+c) ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Alternating series of compositions of triangular numbers I'm modeling a process which involves a subset $S$ of a large number $n_A$ of objects - call them balls. Each time I add a ball to $S$, it may dislodge another ball with probability proportional to the fraction of the total balls already in $S$.
I've gotten as f... | Note that
$$\sum_{a=1}^{k-4}\sum_{b=1}^{a}\sum_{c=1}^{b}\sum_{d=1}^{c}1=
\sum_{a=1}^{k-4}\sum_{b=1}^{a}\sum_{c=1}^{b}\binom c1=\sum_{a=1}^{k-4}\sum_{b=1}^{a}\binom {b+1}2=\sum_{a=1}^{k-4}\binom {a+2}3=\binom {k-1}4$$
and so on.
Hence
$$\begin{align}
\langle n_k\rangle
&=k-
\frac{1}{n_{A}}\sum_{a=1}^{k-1}1+
\frac{1}{{n_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find smallest k for which the inequality holds The smallest positive number $K$ for which the inequality
$|\sin^2 x - \sin^2 y| \le K|x-y|$ holds for all $x,y$ is
| Hint:
Here is sufficient to find the maximum absolute value of the derivative. For this function $2\sin x \cos x$ is $1$. This means that from Lagrange's theorem implies that $K = 1$ is suitable. A lower value will not work, if you take the number of $x, y$ is close to the point $\frac \pi{4}$, where the derivative is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find all the solutions of the congruence $12^x \equiv 17 \bmod 25$
I need to find all the solutions of the congruence $12^x \equiv 17 \bmod 25$.
I don't really have an idea how to approach this..
I tried to write it as:
$12^x \equiv 17 \mod 25$ $\Leftrightarrow$ $(3*2*2)^x \equiv -8 \mod 25 $ $\Leftrightarrow$ $ 3^x... | As said in the comments, you can ultimately reduce to check a finite number of cases. I'll illustrate a standard method which often lessens the number of cases to check. The idea is: since a congruence $\pmod{p^m}$ is also a congruence $\pmod{p^k}$ for each $k<m$, you need not consider the cases for which it fails $\pm... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$ for $a, b, c > 0$ Prove for $a, b, c > 0$ that
$$a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$$
Could you give me some hints on this?
I thought that Jensen's inequality might be of use ... | If you look at C-S for just two terms and its use here becomes more obvious
$$ (a_1b_1 + a_2b_2)^2 \le (a_1^2+a_2^2)(b_1^2+b_2^2) $$
Now look at inequality we want to prove. Square both sides and examine how the terms match with our C-S
$$(a\sqrt{b+c} + b\sqrt{a+c} + c\sqrt{a+b})^2 \le 2(a+b+c)(ab+ac+bc)$$
We can see h... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
$x\cos(x)=-\frac{1}{2} \sin(x) + 2\sum_{n=2}^{\infty} \frac{(-1)^n n \sin(nx)}{n^2-1}$ for $x\in (-\pi,\pi)$ I am trying to establish the following $x\cos(x)=-\frac{1}{2} \sin(x) + 2\sum_{n=2}^{\infty} \frac{(-1)^n n \sin(nx)}{n^2-1}$ for $x\in (-\pi,\pi)$
The right sight looks the the Fourier expansion of the left han... | The function $x\cos(x)$ is an odd function on $(-\pi,\pi)$, which means that its Fourier series on $(-\pi,\pi)$ is a sin series. The function $x\cos(x)|_{x=-\pi}=\pi$ and $x\cos(x)|_{x=\pi}=-\pi$. So the Fourier series converges to $0$ at $x=\pm \pi$; otherwise it converges to $x\cos(x)$ on $(-\pi,\pi)$. Using trig ide... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\int_0^1\frac{1+x^8}{1+x^{10}}dx=\frac{\pi}{\phi^5-8}$ Prove that,
$$
\int_0^1\frac{1+x^8}{1+x^{10}}dx=\frac{\pi}{\phi^5-8}
$$
What kind of subsititution should be used to solve this integral
Another integral that give the same answer but with a different limit and same denominator?
$$
\int_0^\infty \frac{1}... | If we make the substitution $t=\frac1x$ and later $x^{10}=u$, then we find that
$$\begin{align}\int_0^1\frac{1+x^8}{1+x^{10}}dx&=\int_{\infty}^1\frac{1+t^{-8}}{1+t^{-10}}\frac{(-dt)}{t^2}=\int_1^{\infty}\frac{t^8+1}{t^{10}+1}dt\\
&=\frac12\int_0^{\infty}\frac{1+x^8}{1+x^{10}}dx=\frac12\int_0^{\infty}\frac{1+u^{\frac45}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the sum of this series: $1 + \frac{1}{5}x + \frac{1 \times 6}{5 \times 10}x^2 +\cdots$? Say I have a series like the following;
$$1 + \frac{1}{5}x + \frac{1 \times 6}{5 \times 10}x^2 + \frac{1 \times 6 \times 11}{5 \times 10 \times 15}x^3 + \cdots.$$
How do I find the sum of this?
I'm trying to find an expres... | As @Jennifer answered, the general term seems to be
$$
\frac{\prod_{i=0}^{n-1} (5i+1) }{\prod_{i=0}^{n-1} 5(i+1)}=\frac{\Gamma\left(\frac15+n\right)}{\Gamma\left(\frac15\right)\Gamma\left(n+1\right)}=\frac{\left(\frac15\right)_n}{n!} \tag1
$$ where $(x)_n=x(x+1)\cdots (x+n-1)$, then one may recall the generalized bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the limit $\lim\limits_{n\rightarrow \infty}(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\dots+\frac{1}{n^{3}-n})$ $$\displaystyle\lim_{n\rightarrow \infty} \left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\dots+\frac{1}{n^{3}-n}\right)$$
I am not able to find any technique to proceed. It migh... | Since:
$$
\frac{1}{2}\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)=\frac{1}{n^3-n}
$$
We have:
$$
\sum_{n=2}^M\frac{1}{n^3-n}=\sum_{n=2}^M\frac{1}{2}\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)=\frac{1}{4}-\frac{1}{M(M+1)}
$$
So the limit is $\frac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that the sum of the squares of two odd integers cannot be the square of an integer. Prove that the sum of the squares of two odd integers cannot be the square of an integer.
My method:
Assume to the contrary that the sum of the squares of two odd integers can be the square of an integer. Suppose that $x, y, z \in... | You had a great start but you should not make this last factorisation.
$$4(m^2 + n^2) + 4(m + n) + 2=4(m^2+n^2+m+n)+2\equiv 2 \pmod 4$$
But a square can't be equal to $2 \pmod 4$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
What is the inverse of this binary, bijective function? The following function $\operatorname{encode}$ is a binary, bijective function:
$$\operatorname{encode}(x,y) := \binom{x+y+1}{2} + x = z$$
This function comes up in the context of theoretical computer science in order to prove that LOOP-computable functions are pr... | As @AndréNicolas correctly noted, this encoding function is the Cantor pairing function. Given
$$\operatorname{encode}(x,y) = \binom{x+y+1}{2} + x = \frac{1}{2} (x+y)(x+y+1) + x = z$$
we introduce additional variables to help derive the $\operatorname{decode}$ functions:
\begin{align}
w &:= x + y \\
t &:= \binom{x+y+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
sum of binomial series How do I solve sum of binomial series which is as follows:
$$\frac{1}{3}\sum^{\infty}_{x=0}\begin{pmatrix}x\\y\end{pmatrix}\frac{1}{3}^x$$
I think it would be pretty easy to sum from $y=0$ to $x$, but I have no idea how to sum from $x=0$ to $x=\infty$.
Thanks,
| Remember binomial theorem:
$$ (p+q)^x = \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} q^y p^{x-y} $$
With $ p = 1/3 $:
$$ (1/3+q)^x = \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} q^y \left( \frac{1}{3} \right)^{x-y} = \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} 3^y q^y \left( \frac{1}{3} \right)^{x} $$
Now summing over $ x $ on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove using factorials that ${n\choose k}+2{n\choose k+1}+{n\choose k+2}={n+2\choose k+2}$ Prove using factorials that ${n\choose k}+2{n\choose k+1}+{n\choose k+2}={n+2\choose k+2}$
I think I'm having a bit of algebra problem with this proof. Here is my work thus far:$$\frac{n!}{k!(n-k)!}+2\frac{n!}{(k+1)!(n-k-1)!}+\fr... | Write down the LHS as
$${n \choose k} + {n \choose k+1} + {n \choose k+1} + {n \choose k+2}.$$
Now expand the first two terms to get ${n+1\choose k+1}$. The summation of the second two terms evaluates to ${n+1\choose k+2}$. The addition of the resulted two terms is the RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate the double integral $\iint_D\sqrt{4-x^2-y^2}$ bounded by semi-circle I would appreciate it if someone can help me solve this question, as I'm struggling to get its answer.
Q: Evaluate the double integral $$\iint_D\sqrt{4-x^2-y^2}dxdy$$ bounded by semi-circle $$x^2+y^2=2x$$ and lying in first quadrant
Thanks
| Using polar coordinates, $x^2+y^2=2x$ gives $r^2=2r\cos\theta$ so $r=2\cos\theta$.
For the part of the region inside the circle which is also in the first quadrant, we get
$\displaystyle\int_0^{\frac{\pi}{2}}\int_0^{2\cos\theta}\sqrt{4-r^2}\; r\;drd\theta=\int_0^{\frac{\pi}{2}}\left[-\frac{1}{3}(4-r^2)^{\frac{3}{2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, ... | In general if $f$ is a matrix function and $J$ a Jordan block with eigenvalue $\lambda_{0}$ then
\begin{equation}
f(J)=\left(\begin{array}{ccccc}
f(\lambda_{0}) & \frac{f'(\lambda_{0})}{1!} & \frac{f''(\lambda_{0})}{2!} & \ldots & \frac{f^{(n-1)}(\lambda_{0})}{(n-1)!}\\
0 & f(\lambda_{0}) & \frac{f'(\lambda_{0})}{1!} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 4
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.